The ratio test says that, if we have the positive series
$$\sum_{n=1}^{\infty} a_n,$$
such that $\lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = L$, then
(1) if $L < 1$, then $\sum_{n=1}^{\infty}a_n$ is absolutely convergent;
(2) if $L > 1$, then $\sum_{n=1}^{\infty}a_n$ is divergent;
(3) if $L = 1$, then the ratio test gives no information.
I want to understand the reasons behind why and how this works, rather than just memorising formulae.
I have attempted to reason about this theorem myself. It can be seen that we're taking the ratio between the second term in the series, $a_{n+1}$, and the first term in the series, $a_n$. This is exactly how one finds the common ratio in a geometric sequence ($r = \dfrac{a_{n+1}}{a_n} )$. We then take the limit of this ratio, which I assume is to find the relative rate of change between $a_{n+1}$ and $a_n$. In which case, it is more effective to take the absolute value of the limit, $\lim_{n \to \infty} \begin{vmatrix}{ \dfrac{a_{n+1}}{a_n} }\end{vmatrix} = L$. This seems analagous to how the limit comparison test theorem works. Therefore, I presume that, if $L < 1$, then this implies that $a_n$ has a greater rate of change than $a_{n+1}$, which implies that each successive term in the series is getting smaller, and so as we go to infinity, each successive term is converging towards $0$. As such, the series should converge to some value. Analogously, I presume that, if $L > 1$, then this implies that $a_{n+1}$ has a greater rate of change than $a_n$, which implies that each successive term in the series is getting larger, and so as we go to infinity, the terms diverge towards infinity.
Is this reasoning correct? If anything is incorrect, then please clarify why it is incorrect, and what the correct reasoning is.
You have some ideas correct, but remember that $a_n \rightarrow 0$ does not imply that $\sum a_n$ converges. So we need something more than that. However, for the case when $L>1$, your reasoning is correct.
For the case $L<1$, we know that there exists some $r < 1$ and $N \in \mathbb{N}$ such that $$\frac{a_{n+1}}{a_n} \leq r < 1,$$ whenever $n \geq N$.
Therefore,
$$ a_{N+1} \leq ra_N, $$ $$ a_{N+2} \leq ra_{N+1} \leq r^2a_N $$ $$ a_{N+3} \leq ra_{N+2} \leq r^3a_N, $$ and in general $$ a_{N+n} \leq r^na_N. $$
Therefore, at least eventually, we can compare the series from above with a convergent geometric series $\sum a_N r^n$, which implies the convergence of the original series $\sum a_n$.